3.523 \(\int \frac{1}{3+5 \sec (c+d x)} \, dx\)

Optimal. Leaf size=31 \[ \frac{5 \tan ^{-1}\left (\frac{\sin (c+d x)}{\cos (c+d x)+3}\right )}{6 d}-\frac{x}{12} \]

[Out]

-x/12 + (5*ArcTan[Sin[c + d*x]/(3 + Cos[c + d*x])])/(6*d)

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Rubi [A]  time = 0.0308961, antiderivative size = 31, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {3783, 2657} \[ \frac{5 \tan ^{-1}\left (\frac{\sin (c+d x)}{\cos (c+d x)+3}\right )}{6 d}-\frac{x}{12} \]

Antiderivative was successfully verified.

[In]

Int[(3 + 5*Sec[c + d*x])^(-1),x]

[Out]

-x/12 + (5*ArcTan[Sin[c + d*x]/(3 + Cos[c + d*x])])/(6*d)

Rule 3783

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(-1), x_Symbol] :> Simp[x/a, x] - Dist[1/a, Int[1/(1 + (a*Sin[c + d
*x])/b), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0]

Rule 2657

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{q = Rt[a^2 - b^2, 2]}, Simp[x/q, x] + Simp
[(2*ArcTan[(b*Cos[c + d*x])/(a + q + b*Sin[c + d*x])])/(d*q), x]] /; FreeQ[{a, b, c, d}, x] && GtQ[a^2 - b^2,
0] && PosQ[a]

Rubi steps

\begin{align*} \int \frac{1}{3+5 \sec (c+d x)} \, dx &=\frac{x}{3}-\frac{1}{3} \int \frac{1}{1+\frac{3}{5} \cos (c+d x)} \, dx\\ &=-\frac{x}{12}+\frac{5 \tan ^{-1}\left (\frac{\sin (c+d x)}{3+\cos (c+d x)}\right )}{6 d}\\ \end{align*}

Mathematica [A]  time = 0.0555576, size = 30, normalized size = 0.97 \[ \frac{2 (c+d x)+5 \tan ^{-1}\left (2 \cot \left (\frac{1}{2} (c+d x)\right )\right )}{6 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(3 + 5*Sec[c + d*x])^(-1),x]

[Out]

(2*(c + d*x) + 5*ArcTan[2*Cot[(c + d*x)/2]])/(6*d)

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Maple [A]  time = 0.039, size = 34, normalized size = 1.1 \begin{align*}{\frac{2}{3\,d}\arctan \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) }-{\frac{5}{6\,d}\arctan \left ({\frac{1}{2}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) } \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(3+5*sec(d*x+c)),x)

[Out]

2/3/d*arctan(tan(1/2*d*x+1/2*c))-5/6/d*arctan(1/2*tan(1/2*d*x+1/2*c))

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Maxima [A]  time = 1.77395, size = 63, normalized size = 2.03 \begin{align*} \frac{4 \, \arctan \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right ) - 5 \, \arctan \left (\frac{\sin \left (d x + c\right )}{2 \,{\left (\cos \left (d x + c\right ) + 1\right )}}\right )}{6 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(3+5*sec(d*x+c)),x, algorithm="maxima")

[Out]

1/6*(4*arctan(sin(d*x + c)/(cos(d*x + c) + 1)) - 5*arctan(1/2*sin(d*x + c)/(cos(d*x + c) + 1)))/d

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Fricas [A]  time = 1.6259, size = 89, normalized size = 2.87 \begin{align*} \frac{4 \, d x + 5 \, \arctan \left (\frac{5 \, \cos \left (d x + c\right ) + 3}{4 \, \sin \left (d x + c\right )}\right )}{12 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(3+5*sec(d*x+c)),x, algorithm="fricas")

[Out]

1/12*(4*d*x + 5*arctan(1/4*(5*cos(d*x + c) + 3)/sin(d*x + c)))/d

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{5 \sec{\left (c + d x \right )} + 3}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(3+5*sec(d*x+c)),x)

[Out]

Integral(1/(5*sec(c + d*x) + 3), x)

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Giac [A]  time = 1.24765, size = 41, normalized size = 1.32 \begin{align*} -\frac{d x + c - 10 \, \arctan \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 3}\right )}{12 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(3+5*sec(d*x+c)),x, algorithm="giac")

[Out]

-1/12*(d*x + c - 10*arctan(sin(d*x + c)/(cos(d*x + c) + 3)))/d